Question #e32b6

2 Answers
Sep 15, 2015

I_(3(aq))^(-) + 2S_2O_(3(aq))^(2–) -> 3 I_((aq))^(–) + S_4O_(6(aq))^(2–)

Explanation:

  • Oxidation:

2S_2O_(3(aq))^(2–) -> S_4O_(6(aq))^(2–) + 2e^-

  • Reduction: I_(3(aq))^(–) + 2e^(-) -> 3 I_((aq))^(–)

the overall reaction then:

  • Redox:

I_(3(aq))^(–) + 2S_2O_(3(aq))^(2–) -> 3 I_((aq))^(–) + S_4O_(6(aq))^(2–)

From the balanced equation, we can say that:

(1)/"1" * n_(I_3^(–)) = (1)/"2" * n_(S_2O_3^(2–)).

where the denominators 1 and 2 were taking from the corresponding coefficients of I_3^– and S_2O_3^(2–) respectively.

Now,

[I_3^–] . V = (1)/"2" * [S_2O_3^(2–)] * V^'

(since n = C * V), where

V^' = "10.50 mL " and " "V = "15.00 mL"

Therefore,

[I_3^–] = (1)/"2" * [S_2O_3^(2–)] * (V')/"V"

[I_3^–] = (1)/"2" * "0.0500M" * (10.50 cancel("mL"))/(15.00 cancel("mL")) = "0.0175M"

(rounded to 3 significant figures)

I would like to suggest the following video, that deals with a similar example in terms of stoichiometry of a redox reaction.

Sep 15, 2015

(a)" "I_(3(aq))^(-)+2S_2O_(3(aq))^(2-)rarrS_4O_(6(aq))^(2-)+3I_((aq))^-

(b)" "0.0175"mol/l"

Explanation:

Thiosulfate ions give out electrons:

2S_2O_(3(aq))^(2-)rarrS_4O_(6(aq))^(2-)+2e

These are taken in by I_3^- which are reduced:

I_(3(aq))^(-)+2erarr3I_((aq))^(-)

Adding both sides gives:

I_(3(aq))^(-)+2S_2O_(3(aq))^(2-)rarrS_4O_(6(aq))^(2-)+3I_((aq))^-

(b).

c=n/v

n=cxxv

So no. moles S_2O_3^(2-)=0.05xx0.0105=5.25xx10^(-4)

From the equation we can see that the number of moles of I_3^- must be half of this:

no. moles I_3^(-)=(5.25xx10^(-4))/(2)=2.625xx10^(-4)

c=n/v

So [I_(3(aq))^(-)]=(2.625xx10^(-4))/(0.015)=0.0175"mol/l"